Notes on Lewin Appendix

Appendix A: Sobolev Spaces

Definition

Definition of ${𝑊}^{𝑘,𝑝}$ is given, and the Meyer-Serrin theorem $𝐻=𝑊$ is stated without proof.

Sobolev spaces on the interval

Considering $𝑓\in {𝐿}^1$ with distributional derivative ${𝑓}^{\prime }\in {𝐿}^1$. Then $𝑔(𝑥)≔{\int }_0^{𝑥}𝑓(𝑡)d𝑡$ is continuous¹ on $[0,1]$ and ${𝑔}^{\prime }={𝑓}^{\prime }$ in ${\mathrm{𝒟︀}}^{\prime }$, so $𝑓$ is a.e. equal to a continuous function on $[0,1]$, and we identify $𝑓$ with this continuous representative. This lets us write $𝑓(𝑥)=𝑓(0)+{\int }_0^{𝑥}{𝑓}^{\prime }(𝑡)d𝑡$. This implies that $𝑓(𝑥)=𝑓(𝑦)+{\int }_{𝑦}^{𝑥}{𝑓}^{\prime }(𝑡)d𝑡$, which after integrating, implies the estimate

If $𝑓\in {𝐻}^1$, we can use this estimate on $|𝑓{|}^2\in {𝐿}^1$ with the identity

to get the variant inequality ${max}_{𝑥\in [0,1]}|𝑓(𝑥){|}^2\le \Vert 𝑓{\Vert }_{{𝐿}^2}^2+2\Vert 𝑓{\Vert }_{{𝐿}^2}\Vert {𝑓}^{\prime }{\Vert }_{{𝐿}^2},$ or after the subadditivity of the square root,

Exercise A.3 - Density of smooth functions

Let $𝑓\in {𝐻}^1$. (1) Construct ${𝑓}_{𝑛}$ in ${𝐶}^{\infty }$ with ${𝑓}_{𝑛}\to 𝑓$ in ${𝐻}^1$. (2) If we further assume that $𝑓(0)=𝑓(1)=0$, then do the same with ${𝑓}_{𝑛}\in {𝐶}_{𝑐}^{\infty }$. (3) Finally, extend this result to ${𝐻}^{𝑘}$. Solutions

1. Extend $𝑓$ by zero outside $[0,1]$ to get a function in ${𝐻}^1(ℝ)$. Then convolve with a mollifier ${𝜑}_{𝑛}(𝑥)=𝑛𝜑(𝑛𝑥)$ for $𝜑\in {𝐶}_{𝑐}^{\infty }([-1,1])$ with $𝜑\ge 0,\int 𝜑=1$ to get the desired sequence of smooth functions - one gets

   $$\Vert {𝜑}_{𝑛}\ast 𝑓-𝑓{\Vert }_{𝑝}\le \int |{𝜑}_{𝑛}(𝑦)|\Vert 𝑓(.-𝑦)-𝑓{\Vert }_{𝑝}d𝑦\to 0$$

   and similarly with the derivative, whence the result.

2. We use a cutoff $𝜒\in {𝐶}_{𝑐}^{\infty }\left(\left[0,\frac{2}{3}\right]\right)$ with $𝜒{|}_{\left[0,\frac{1}{3}\right]}=1$ to define ${𝑓}^1\equiv 𝜒𝑓$ and ${𝑓}^2=(1-𝜒)𝑓$. Then $𝑓={𝑓}^1+{𝑓}^2$. We focus on ${𝑓}^1$; how to treat ${𝑓}^2$ will be clear. We extend ${𝑓}^1$ to a function on $ℝ$ that is zero for $𝑥>\frac{2}{3}$ or $𝑥<0$. By the given boundary conditions, ${𝑓}^1\in {𝐻}^1$. Indeed, the derivative of the extension by zero will have the term $𝑓(0){𝛿}_0\notin {𝐿}^2$ unless $𝑓(0)=0$. We next need ${𝑓}_{𝑛}^1\in {𝐶}_{𝑐}^{\infty }$ that converge in ${𝐻}^1$ to ${𝑓}^1$. To do this we modify the convolution formula by setting

   $${𝑓}_{𝑛}^1(𝑥)={\int }_{{𝔹}_{\frac{1}{𝑛}}(0)}{𝜑}_{𝑛}(𝑦){𝑓}^1\left(𝑥-𝑦-\frac{2}{𝑛}\right)d𝑦={𝜑}_{𝑛}\ast {𝑓}^1\left(𝑥-\frac{2}{𝑛}\right).$$

   Then ${𝑓}_{𝑛}\in {𝐶}^{\infty }$ and is zero if $𝑥\le \frac{1}{𝑛}$ or $𝑥>\frac{2}{3}-\frac{1}{𝑛}$, and for the convergence we have

   $$\Vert {𝑓}_{𝑛}^1-{𝑓}^1{\Vert }_{𝑝}\le {\Vert {𝜑}_{𝑛}\ast {𝑓}^1\left(𝑥-\frac{2}{𝑛}\right)-{𝜑}_{𝑛}\ast {𝑓}^1\Vert }_{𝑝}+\Vert {𝜑}_{𝑛}\ast {𝑓}^1-{𝑓}^1{\Vert }_{𝑝}$$

   The first term converges by Young’s inequality for convolutions and the second by good kernel properties. One can then repeat this for ${({𝑓}^1)}^{\prime }$ and then symmetrically for ${𝑓}^2$ to get the result.

3. Exactly the same method works for ${𝐻}^{𝑘}$.

Elliptic Regularity on $(0,1)$

This is the statement that if $𝑓,{𝑓}^{″}\in {𝐿}^2$ then ${𝑓}^{\prime }\in {𝐶}^0$ with the estimate ${max}_{𝑥\in [0,1]}|{𝑓}^{\prime }(𝑥)|\le 𝐶(\Vert 𝑓{\Vert }_{{𝐿}^2}+\Vert {𝑓}^{″}{\Vert }_{{𝐿}^2})$. We recap the proof given on $[0,1]$. Put $𝑤(𝑦)=𝑦(1-𝑦)$. First, we know that ${𝑓}^{″}\in {𝐿}^2\subset {𝐿}^1$ implies that $𝑓\in {𝐶}^{1,\frac{1}{2}}$. Then we write ${𝑓}^{\prime }(𝑦)={𝑓}^{\prime }(0)+{\int }_0^{𝑦}{𝑓}^{″}(𝑡)d𝑡$, and integrate it against $𝑤$:

Integrating by parts we get ${\int }_0^1𝑤(𝑦){𝑓}^{\prime }(𝑦)d𝑦={𝑤(𝑦)𝑓(𝑦)|}_0^1-{\int }_0^1{𝑤}^{\prime }(𝑦)𝑓(𝑦)d𝑦$. The boundary term vanishes since $𝑤(0)=𝑤(1)=0$. For the second term on the right, we write

This gives the identity

This leads to a bound of ${𝑓}^{\prime }$ using only $𝑓$ and ${𝑓}^{″}$. We see that $𝑤$ was introduced so that we could use $𝑤(0)=𝑤(1)=0$ (and ${\int }_0^1𝑤\ne 0$).

Exercise A.4 - Half line $\left[0,\infty \right)$

1. Show that ${𝐻}^1\hookrightarrow ︎{𝐶}_0^0$
2. Show that ${𝐶}^{\infty }$ is dense in ${𝐻}^1$, and that ${𝐻}_0^1$ is the space of functions in ${𝐻}^1$ that vanish at zero.
3. Prove the elliptic estimate (bound ${𝑓}^{\prime }$ by $𝑓$ and ${𝑓}^{″}$.)

Solutions:

1. All the estimates trivially carry over as we can just restrict $𝑓$ to an interval and use the previous estimates. The only new thing we need to show is that $𝑓(𝑥)\to 0$ as $𝑥\to \infty$. But this follows from the fact that ${\sum }_{𝑛}\Vert 𝑓{\Vert }_{{𝐻}^1([𝑛,𝑛+1])}=\Vert 𝑓{\Vert }_{{𝐻}^1(\left[0,\infty \right))}<\infty$ and the ${𝐿}^{\infty }$ estimate

   $$\underset{𝑥\in [𝑛,𝑛+1]}{sup}|𝑓(𝑥)|\lesssim \Vert 𝑓{\Vert }_{{𝐻}^1([𝑛,𝑛+1])}\to 0.$$
2. Same proof works.
3. The result follows immediately by restricting $𝑓$ to intervals. But we could also redo the proof with some weight $𝑤\in {𝐿}^1$ such that ${𝑤}^{\prime }\in {𝐿}^{{𝑝}^{\prime }}$ and ${lim}_{|𝑥|\to \infty }𝑤(𝑥)=0$.

The Sobolev Space ${𝐻}^{𝑠}({ℝ}^{𝑑})$

The usual norm for $𝑠\in \mathbb{N}$ is $\Vert 𝑓{\Vert }_{{𝐻}^{𝑠}}^2={\sum }_{|𝛼|\le 𝑘}\Vert {𝐷}^{𝛼}𝑓{\Vert }_{{𝐿}^2}^2$. Note that we have the trivial inequality $|{𝜉}^{𝛼}{|}^2\le |𝜉{|}^{2|𝛼|}\le {(1+|𝜉{|}^2)}^{|𝛼|}$, and also by multinomial theorem that $|𝜉{|}^{2𝑠}={({𝜉}_1^2+\dots +{𝜉}_{𝑑}^2)}^{𝑠}={\sum }_{|𝛽|=𝑠}|{𝜉}^{𝛽}{|}^2$. This implies by Fourier transform the equivalence of norms. This is also equivalent to the ‘elliptic estimate’ $\Vert 𝑓{\Vert }_{{𝐻}^{𝑠}}\le 𝐶\left(\Vert 𝑓{\Vert }_{{𝐿}^2}+\Vert {\mathrm{\Delta }}^{\frac{𝑠}{2}}𝑓{\Vert }_{{𝐿}^2}\right)$.

Trace, Lifting and Extension on a Bounded Open Set

First the half-space $\mathrm{\Omega }=\{𝑥=({𝑥}_1,\dots ,{𝑥}_{𝑑}):{𝑥}_{𝑑}>0\}$ is treated. For general spaces the ${𝐿}^{𝑝}$ and ${𝐻}^{𝑠}$ spaces are defined by local charts and partitions of unity, so the half-space case is the main one to understand.

Recall that if $\int |𝑓|<\infty$ then $|𝑓|<\infty$ a.e. So if $𝑓\in {𝐻}^1(\mathrm{\Omega })$, then for a.e. ${𝑥}^{\prime }=({𝑥}_1,\dots ,{𝑥}_{𝑑-1})$, the function ${𝑔}_{{𝑥}^{\prime }}({𝑥}_{𝑑})≔𝑓({𝑥}^{\prime },{𝑥}_{𝑑})$ belongs to ${𝐻}^1((0,\infty ))$ by Fubini’s theorem -

So for a.e. ${𝑥}^{\prime }$, ${𝑔}_{{𝑥}^{\prime }}$ coincides a.e. in ${𝑥}_{𝑑}$ with a continuous function on $\left[0,\infty \right)$ that vanishes at infinity by earlier exercise. We can then define the trace of $𝑓$ at ${𝑥}_{𝑑}=0$ to be this continuous representative, which is well-defined for a.e. ${𝑥}^{\prime }$. This gives us a well-defined and bounded (again, by earlier exercise) trace operator $𝑇:{𝐻}^1(\mathrm{\Omega })\to {𝐿}^2({ℝ}^{𝑑-1})$.

For ${𝐻}^{𝑘}$, $𝑘>1$, the trace map is $𝑇:{𝐻}^{𝑘}(\mathrm{\Omega })\to {𝐿}^2{(𝜕\mathrm{\Omega })}^{𝑘}$, mapping

with the bound

Stated without proof:

• $𝑓\in {𝐻}_0^{𝑘}(\mathrm{\Omega })$ iff $𝑓\in {𝐻}^{𝑘}$ and $𝑇𝑓=0$.
• Theorem A.10 (“smooth lifting”): Given continuously differentiable boundary values we can find $𝑓\in {𝐻}^{𝑘}(\mathrm{\Omega })$ such that $𝑇𝑓$ equals the given boundary values.
• Fractional Sobolev space defined by weighted integral of difference quotient ${𝐻}^{𝑠}(\mathrm{\Omega })$ to state that this space is the image of ${𝐻}^{𝑠}({ℝ}^{𝑑})$ under the restriction map, and that the trace map is (or rather can be) bounded from ${𝐻}^{𝑠}(\mathrm{\Omega })$ to ${𝐻}^{𝑠-\frac{1}{2}}(𝜕\mathrm{\Omega })$.

Sobolev Embedding and Rellich-Kondrachov Compactness

Interesting version of Sobolev embedding is proven- let $𝑑\ge 1$ and $0<𝑠<\frac{𝑑}{2}$. For any $𝑓\in {𝐿}_{\text{loc}({ℝ}^{𝑑})}^1\cap {\mathrm{𝒮︀}}^{\prime }$ such that the superlevel sets $\{𝑓>𝑀\}$ have finite measure for all $𝑀>0$, then

where ${𝑝}^{\ast }$ is the number such that $\frac{1}{{𝑝}^{\ast }}=\frac{1}{2}-\frac{𝑠}{𝑑}$. There is a version for Bessel potential spaces with $𝑝\ne 2$, but the proof given only extends to $𝑝\ge 2$ with the Hausdorff-Young inequality.

Then a version o the RK Compactness theorem is given - if $\mathrm{\Omega }$ is a bounded open set, then the embedding ${𝐻}^{𝑠}(\mathrm{\Omega })\hookrightarrow ︎{𝐿}^2(\mathrm{\Omega })$ is compact for any $𝑠>0$. (Here ${𝐻}^{𝑠}(\mathrm{\Omega })$ is the space of restrictions of functions in ${𝐻}^{𝑠}({ℝ}^{𝑑})$.) The proof uses the Fourier transform and the fact that the Fourier transform of a compactly supported function is smooth and decays at infinity. In particular this lemma is used- Let $𝐹\in {𝐿}^{\infty ({ℝ}^{𝑑})}$ decay at infinity, and let $𝐺\in {𝐿}^1({ℝ}^{𝑑})$. Then $𝑔\mapsto (𝐺\ast 𝑔)𝐹$ is compact on ${𝐿}^2({ℝ}^{𝑑})$.

Here it is used that compactness means upgrading from from weak convergence to strong convergence. This is equivalent to the usual extraction of convergent subsequences on reflexive spaces (so not ${𝐿}^1$). In one direction it is used that every subsequence has a further subsequence that converges, and in the other direction Banach-Alaoglu is used to get a weakly convergent subsequence, and then the compactness is used to upgrade to strong convergence.

Evans has a slightly sharper version of the theorem, namely that ${𝑊}^{𝑠,𝑝}$ is compactly embedded in ${𝐿}^{𝑞}$ for $1\le 𝑞<{𝑝}^{\ast }$.

Elliptic Regularity on bounded domains

Without conditions on $𝑓$ at the boundary, the result is false. The example given is the Fundamental solution of the Laplacian translated to the boundary, which is in ${𝐿}^2$ and has $\mathrm{\Delta }𝑓\in {𝐿}^2$, but is not in ${𝐻}^1$ if the boundary supports a cone. The elliptic regularity result holds under Robin boundary conditions.